Answer:
0.150 g
Explanation:
First, we will write the reduction half-reaction to produce Ga(s)
Ga³⁺(aq) + 3 e⁻ → Ga(s)
We can establish the following relations:
The mass of Ga(s) deposited when a current of 0.520 A flows for 20.0 min is:
[tex]20.0min.\frac{60s}{1min} .\frac{0.520c}{s} .\frac{1mol/e^{-} }{96468c} .\frac{1molGa}{3mol/e^{-}} .\frac{69.72gGa}{1molGa} =0.150gGa[/tex]